The Stacks project

Lemma 28.3.4. A scheme $X$ is integral if and only if it is reduced and irreducible.

Proof. If $X$ is irreducible, then every affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ is irreducible. If $X$ is reduced, then $R$ is reduced, by Lemma 28.3.2 above. Hence $R$ is reduced and $(0)$ is a prime ideal, i.e., $R$ is an integral domain.

If $X$ is integral, then for every nonempty affine open $\mathop{\mathrm{Spec}}(R) = U \subset X$ the ring $R$ is reduced and hence $X$ is reduced by Lemma 28.3.2. Moreover, every nonempty affine open is irreducible. Hence $X$ is irreducible, see Lemma 28.3.3. $\square$

Comments (0)

There are also:

  • 3 comment(s) on Section 28.3: Integral, irreducible, and reduced schemes

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01ON. Beware of the difference between the letter 'O' and the digit '0'.



View Lemma 28.3.4 as pdf